1. Language acquisition
(4:30)

2. Language modeling: Smoothing
David Kauchak
CS159 – Spring 2019
some slides adapted from Jason Eisner

3. Admin Assignment 2 out bigram language modeling Java
Can work with partners
Anyone looking for a partner?
2a: Due this Friday
2b: Due next Friday
Style/commenting (JavaDoc)
Some advice
Start now!
Spend 1-2 hours working out an example by hand (you can check your answers with me)
HashMap

4. Admin Our first quiz (when?) In-class (~30 min.) Topics
corpus analysis
regular expressions
probability
language modeling
Open book/notes
we’ll try it out for this one
better to assume closed book (30 minutes goes by fast!)
10% of your grade

5. Admin
Lab next class Meet in Edmunds 105

6. Today
smoothing techniques

7. Today
Take home ideas:
Key idea of smoothing is to redistribute the probability to handle less seen (or never seen) events
Still must always maintain a true probability distribution
Lots of ways of smoothing data
Should take into account features in your data!

8. Smoothing
What if our test set contains the following sentence, but one of the trigrams never occurred in our training data?
P(I think today is a good day to be me) =
P(I | <start> <start>) x
P(think | <start> I) x
If any of these has never been seen before, prob = 0!
P(today| I think) x
P(is| think today) x
P(a| today is) x
P(good| is a) x …

9. Smoothing
P(I think today is a good day to be me) =
P(I | <start> <start>) x
P(think | <start> I) x
These probability estimates may be inaccurate. Smoothing can help reduce some of the noise.
P(today| I think) x
P(is| think today) x
P(a| today is) x
P(good| is a) x …

10. The general smoothing problem
modification
probability
see the abacus 1
1/3 ?
see the abbot
0/3
see the abduct
see the above 2
2/3
see the Abram …
see the zygote
Total 3
3/3

11. Add-lambda smoothing
A large dictionary makes novel events too probable.
add  = 0.01 to all counts
see the abacus 1
1/3
1.01
1.01/203
see the abbot
0/3
0.01
0.01/203
see the abduct
see the above 2
2/3
2.01
2.01/203
see the Abram …
see the zygote
Total 3
3/3
203

12. Add-lambda smoothing How should we pick lambda? 1 1/3 1.01 1.01/203
see the abacus 1
1/3
1.01
1.01/203
see the abbot
0/3
0.01
0.01/203
see the abduct
see the above 2
2/3
2.01
2.01/203
see the Abram …
see the zygote
Total 3
3/3
203

13. Setting smoothing parameters
Idea 1: try many  values & report the one that gets the best results?
Training
Test
Is this fair/appropriate?

14. Setting smoothing parameters
Training
Test
… and report results of that final model on test data.
Now use that  to get smoothed counts from all 100% …
Training
Dev.
collect counts from 80% of the data
pick  that gets best results on 20% … 14
problems? ideas?

15. Vocabulary n-gram language modeling assumes we have a fixed vocabulary
why?
Probability distributions are over finite events!
What happens when we encounter a word not in our vocabulary (Out Of Vocabulary)?
If we don’t do anything, prob = 0
Smoothing doesn’t really help us with this!

16. Vocabulary To make this explicit, smoothing helps us with… 1 1.01 0.01
all entries in our vocabulary
see the abacus 1
1.01
see the abbot
0.01
see the abduct
see the above 2
2.01
see the Abram …
see the zygote

17. Vocabulary and… Vocabulary Counts Smoothed counts
able
about
account
acid
across …
young
zebra 10 1 2 3 …
10.01
1.01
2.01
0.01
3.01 …
some people find this a bit weird, how could we have
How can we have words in our vocabulary we’ve never seen before?

18. Vocabulary Choosing a vocabulary: ideas? Benefits/drawbacks?
Grab a list of English words from somewhere
Use all of the words in your training data
Use some of the words in your training data
for example, all those the occur more than k times
Benefits/drawbacks?
Ideally your vocabulary should represents words you’re likely to see
Too many words: end up washing out your probability estimates (and getting poor estimates)
Too few: lots of out of vocabulary

19. Vocabulary
No matter how you chose your vocabulary, you’re still going to have out of vocabulary (OOV) words
How can we deal with this?
Ignore words we’ve never seen before
Somewhat unsatisfying, though can work depending on the application
Probability is then dependent on how many in vocabulary words are seen in a sentence/text
Use a special symbol for OOV words and estimate the probability of out of vocabulary

20. Out of vocabulary
Add an extra word in your vocabulary to denote OOV (<OOV>, <UNK>)
Replace all words in your training corpus not in the vocabulary with <UNK>
You’ll get bigrams, trigrams, etc with <UNK>
p(<UNK> | “I am”)
p(fast | “I <UNK>”)
During testing, similarly replace all OOV with <UNK>

21. Choosing a vocabulary
A common approach (and the one we’ll use for the assignment):
Replace the first occurrence of each word by <UNK> in a data set
Estimate probabilities normally
Vocabulary then is all words that occurred two or more times
This also discounts all word counts by 1 and gives that probability mass to <UNK>

22. Storing the table How are we storing this table?
Should we store all entries?
see the abacus 1
1/3
1.01
1.01/203
see the abbot
0/3
0.01
0.01/203
see the abduct
see the above 2
2/3
2.01
2.01/203
see the Abram …
see the zygote
Total 3
3/3
203

23. Storing the table Hashtable (e.g. HashMap)
fast retrieval
fairly good memory usage
Only store those entries of things we’ve seen
for example, we don’t store |V|3 trigrams
For trigrams we can:
Store one hashtable with bigrams as keys
Store a hashtable of hashtables (I’m recommending this)

24. Storing the table: add-lambda smoothing
For those we’ve seen before:
Unsmoothed (MLE)
add-lambda smoothing
see the abacus 1
1/3
1.01
1.01/203
see the abbot
0/3
0.01
0.01/203
see the abduct
see the above 2
2/3
2.01
2.01/203
see the Abram …
see the zygote
Total 3
3/3
203
What value do we need here to make sure it stays a probability distribution?

25. Storing the table: add-lambda smoothing
For those we’ve seen before:
Unsmoothed (MLE)
add-lambda smoothing
see the abacus 1
1/3
1.01
1.01/203
see the abbot
0/3
0.01
0.01/203
see the abduct
see the above 2
2/3
2.01
2.01/203
see the Abram …
see the zygote
Total 3
3/3
203
For each word in the vocabulary, we pretend we’ve seen it λ times more (V = vocabulary size).

26. Storing the table: add-lambda smoothing
For those we’ve seen before:
Unseen n-grams: p(z|ab) = ?

27. Problems with frequency based smoothing
The following bigrams have never been seen:
p( X | San )
p( X| ate)
ignores a lot of information!
Which would add-lambda pick as most likely?
Which would you pick?

28. Witten-Bell Discounting
Some words are more likely to be followed by new words
food
apples
bananas
hamburgers
a lot
for two
grapes …
Diego
Francisco
Luis
Jose
Marcos
San
ate

29. Witten-Bell Discounting
Probability mass is shifted around, depending on the context of words
If P(wi | wi-1,…,wi-m) = 0, then the smoothed probability PWB(wi | wi-1,…,wi-m) is higher if the sequence wi-1,…,wi-m occurs with many different words wk

30. Problems with frequency based smoothing
The following trigrams have never been seen:
p( car | see the )
p( zygote | see the )
p( cumquat | see the )
ignores a lot of information!
Which would add-lambda pick as most likely? Witten-Bell?
Which would you pick?

31. Better smoothing approaches
Utilize information in lower-order models
Interpolation
Combine probabilities of lower-order models in some linear combination
Backoff
Often k = 0 (or 1)
Combine the probabilities by “backing off” to lower models only when we don’t have enough information

32. Smoothing: simple interpolation
Trigram is very context specific, very noisy
Unigram is context-independent, smooth
Interpolate Trigram, Bigram, Unigram for best combination
How should we determine λ and μ?

33. Smoothing: finding parameter values
Just like we talked about before, split training data into training and development
Try lots of different values for   on heldout data, pick best
Two approaches for finding these efficiently
EM (expectation maximization)
“Powell search” – see Numerical Recipes in C

34. Backoff models: absolute discounting
Subtract some absolute number from each of the counts (e.g. 0.75)
How will this affect rare words?
How will this affect common words?

35. Backoff models: absolute discounting
Subtract some absolute number from each of the counts (e.g. 0.75)
will have a large effect on low counts (rare words)
will have a small effect on large counts (common words)

36. Backoff models: absolute discounting
What is α(xy)?

37. Backoff models: absolute discounting
trigram model: p(z|xy)
(before discounting)
trigram model p(z|xy)
(after discounting)
bigram model p(z|y)*
(*for z where xyz didn’t occur)
seen trigrams
(xyz occurred)
seen trigrams
(xyz occurred)
(xyz didn’t occur)
unseen words

38. Backoff models: absolute discounting
see the dog 1
see the cat 2
see the banana 4
see the man 1
see the woman 1
see the car 1
p( cat | see the ) = ?
p( puppy | see the ) = ?

39. Backoff models: absolute discounting
see the dog 1
see the cat 2
see the banana 4
see the man 1
see the woman 1
see the car 1
p( cat | see the ) = ?

40. Backoff models: absolute discounting
see the dog 1
see the cat 2
see the banana 4
see the man 1
see the woman 1
see the car 1
p( puppy | see the ) = ?
α(see the) = ?
How much probability mass did we reserve/discount for the bigram model?

41. Backoff models: absolute discounting
see the dog 1
see the cat 2
see the banana 4
see the man 1
see the woman 1
see the car 1
p( puppy | see the ) = ?
α(see the) = ?
# of types starting with “see the” * D
count(“see the”)
for each of the types,
For each of the unique trigrams, we subtracted D/count(“see the”) from the probability distribution

42. Backoff models: absolute discounting
see the dog 1
see the cat 2
see the banana 4
see the man 1
see the woman 1
see the car 1
p( puppy | see the ) = ?
α(see the) = ?
# of types starting with “see the” * D
count(“see the”)
for each of the types,
distribute this probability mass to all bigrams that we are backing off to

43. Calculating α
We have some number of bigrams we’re going to backoff to, i.e. those X where C(see the X) = 0, that is unseen trigrams starting with “see the” When we backoff, for each of these, we’ll be including their probability in the model: P(X | the) α is the normalizing constant so that the sum of these probabilities equals the reserved probability mass

44. Calculating α We can calculate α two ways
Based on those we haven’t seen:
Or, more often, based on those we do see:

45. Calculating α in general: trigrams
p( C | A B )
Calculate the reserved mass
Calculate the sum of the backed off probability. For bigram “A B”:
Calculate α
# of types starting with bigram * D
reserved_mass(bigram--A B) =
count(bigram)
either is fine, in practice the left is easier
1 – the sum of the bigram probabilities of those trigrams that we saw starting with bigram A B

46. Calculating α in general: bigrams
p( B | A )
Calculate the reserved mass
Calculate the sum of the backed off probability. For bigram “A B”:
Calculate α
# of types starting with unigram * D
reserved_mass(unigram--A) =
count(unigram)
either is fine in practice, the left is easier
1 – the sum of the unigram probabilities of those bigrams that we saw starting with word A

47. Calculating backoff models in practice
Store the αs in another table
If it’s a trigram backed off to a bigram, it’s a table keyed by the bigrams
If it’s a bigram backed off to a unigram, it’s a table keyed by the unigrams
Compute the αs during training
After calculating all of the probabilities of seen unigrams/bigrams/trigrams
Go back through and calculate the αs (you should have all of the information you need)
During testing, it should then be easy to apply the backoff model with the αs pre-calculated

48. Backoff models: absolute discounting
the Dow Jones 10
the Dow rose 5
the Dow fell 5
p( jumped | the Dow ) = ?
What is the reserved mass?
# of types starting with “the Dow” * D
count(“the Dow”)

49. Backoff models: absolute discounting
# of types starting with bigram * D
count(bigram)
reserved_mass =
Two nice attributes:
decreases if we’ve seen more bigrams
should be more confident that the unseen trigram is no good
increases if the bigram tends to be followed by lots of other words
will be more likely to see an unseen trigram